1.3 Solutions
35
1.3.4 Force and Torque
1.39 Resolve the force into x- and y-components:
F x = −80 cos 35
◦
+ 60 + 40 cos 45
◦
= 22.75 N
F y = 80 sin 35
◦
+ 0 − 40 sin 45
◦
= 17.6 N
(i) F net =
F 2
x + F 2
y =
(22.75) 2 + (17.6) 2 = 28.76 N
tan θ =
F y
F x
=
17.6
22.75
= 0.7736 → θ = 37.7
◦
The vector F net makes an angle of 37.7 ◦ with the x-axis.
(ii) a =
F net
m
=
28.76 N
3.8 kg
= 7.568 m/s 2
(iii) F 4 of magnitude 28.76 N must be applied in the opposite direction to
F net
1.40 (a) (i) τ = r × F
τ = r F sin θ = (0.4 m)(50 N) sin 90
◦
= 20 N − m
(ii) τ = I α
α =
τ
I
=
20
20
= 1.0 rad/s
2
(iii) ω = ω 0 + αt = 0 + 1 × 3 = 3 rad/s
(iv) ω 2 = ω 2
0 + 2αθ, θ =
3 2 −0
2×1 = 4.5 rad
(b) (i) τ = 0.4 × 50 × sin(90 + 20) = 18.794 N m
(ii) α =
τ
I
=
18.794
20
= 0.9397 rad/s
2
1.41 Force applied to the container F = ma
Frictional force = F r = μ mg
F r = F
μ mg = ma
μ =
a
g
=
1.5
9.8
= 0.153